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Stability of the species $Li_2, Li_2^-$ and $Li_2^+$ increases in the order of :
$Li_2 < Li_2^+ < Li_2^-$
$Li_2^- < Li_2^+ < Li_2$
$Li_2 < Li_2^- < Li_2^+$
$Li_2^- < Li_2 < Li_2^+$
Solution
Correct option
(b) $\mathrm{Li}_{2}-<\mathrm{Li}_{2}^{+}<\mathrm{Li}_{2}$
Explanation:
$\mathrm{Li}_{2}(3+3=6)=\sigma 1 \mathrm{s}^{2}, \quad \hat{\sigma} 1 \mathrm{s}^{2}, \sigma 2 \mathrm{s}^{2}$
Bond order $=\frac{N_{b}-N_{a}}{2}=\frac{4-2}{2}=1$
$\mathrm{Li}_{2}^{+}(3+3-1=5)=\sigma 1 \mathrm{s}^{2}, \dot{\sigma}^{2} 1 s^{2}, \cdot \sigma 2 s^{1}$
Bond order $=\frac{3-2}{2}=\frac{1}{2}=0.5$
Bond order $=\frac{4-3}{2}=\frac{1}{2}=0.5$
Stability order is $\mathrm{Li}_{2}>\mathrm{Li}_{2}^{+}>\mathrm{Li}_{2}^{-}$ (because Li $_{2}^{-}$ have more number of electrons in antibonding orbitals which destabilises the species).